5.2 Op-Amp Linear Circuits & Calculations

5.2 Op-Amp Linear Circuits & Calculations

Operational amplifiers achieve linear, predictable amplification through the deliberate application of negative feedback. By feeding a fraction of the output signal back to the inverting ($-$) input terminal, the extreme and temperature-sensitive open-loop gain ($A_{OL} \approx 10^5\text{ to }10^6$) is traded for precise, highly stable closed-loop amplification established strictly by external passive resistors. Under the EASA Part-66 syllabus, aircraft maintenance engineers must be capable of deriving circuit transfer functions, calculating component values, assessing input/output impedances, and verifying that signal amplitudes do not violate aircraft DC supply rail saturation boundaries.

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The Principle of Negative Feedback

In an open-loop configuration, minor temperature shifts, power supply variations, or minuscule DC offset drifts cause the amplifier output to slam uncontrollably into positive or negative saturation. Negative feedback introduces a feedback network (typically resistors $R_f$ and $R_{in}$) that returns an opposing signal to the inverting terminal:

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The closed-loop gain ($A_{cl}$) of the feedback system is expressed by the classic Black feedback equation:

$$A_{cl} = \frac{A_{OL}}{1 + \beta A_{OL}}$$

Where $\beta$ is the feedback transfer factor ($0 < \beta \le 1$). When the open-loop gain is enormous such that $\beta A_{OL} \gg 1$:

$$A_{cl} \approx \frac{A_{OL}}{\beta A_{OL}} = \frac{1}{\beta}$$

This fundamental result proves that closed-loop gain is completely independent of internal transistor parameters, temperature drift, and supply fluctuations, depending entirely on the ratio of stable, low-tolerance external metal-film resistors.

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The Inverting Amplifier

In the inverting amplifier, the input voltage ($V_{in}$) is applied through an input resistor ($R_{in}$) to the inverting input ($-$), while a feedback resistor ($R_f$) connects the output ($V_{out}$) back to the inverting input. The non-inverting input ($+$) is connected directly to circuit ground ($0\text{ V}$).

                     +------------[ Rf ]------------+
                     |                              |
                     |    |\                        |
  V_in o----[ Rin ]--+----| -\                      |
                          |   \                     |
                     0V --| +  >--------------------+----o V_out
                          |   / 
                          |/    

Mathematical Derivation via Virtual Ground

1. Because the non-inverting terminal is grounded ($V^+ = 0\text{ V}$), the virtual ground principle dictates that the inverting terminal is also constrained to zero volts: $$V^- = V^+ = 0\text{ V}$$
2. Calculate the current flowing into the summing junction from the input source ($I_{in}$): $$I_{in} = \frac{V_{in} - V^-}{R_{in}} = \frac{V_{in} - 0}{R_{in}} = \frac{V_{in}}{R_{in}}$$
3. Calculate the feedback current flowing from the summing junction to the output ($I_f$): $$I_f = \frac{V^- - V_{out}}{R_f} = \frac{0 - V_{out}}{R_f} = -\frac{V_{out}}{R_f}$$
4. Apply Kirchhoff's Current Law (KCL) at the inverting summing node. Because the ideal op-amp possesses infinite input impedance ($Z_{in} = \infty$), zero current enters the inverting terminal pin ($I^- = 0\text{ A}$): $$I_{in} = I_f$$
5. Equating the current expressions: $$\frac{V_{in}}{R_{in}} = -\frac{V_{out}}{R_f} \implies \frac{V_{out}}{V_{in}} = -\frac{R_f}{R_{in}}$$ $$A_v = -\frac{R_f}{R_{in}}$$

Key Inverting Amplifier Characteristics

• Phase Inversion: The negative sign designates a $180^\circ$ electrical phase shift between input and output.
• Input Impedance ($Z_{in}$): Because the right-hand side of $R_{in}$ terminates at the virtual ground node ($0\text{ V}$), the signal source "sees" an input impedance exactly equal to $R_{in}$ ($Z_{in} = R_{in}$). If $R_{in}$ is selected too low, it will load down high-impedance aircraft transducers.
• Output Impedance ($Z_{out}$): Closed-loop negative feedback drives output impedance down to milliohms ($Z_{out} \approx 0\ \Omega$).

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The Non-Inverting Amplifier

In the non-inverting amplifier, the input signal ($V_{in}$) is connected directly to the high-impedance non-inverting input ($+$). Negative feedback is established by a voltage divider network consisting of $R_f$ and $R_1$ connected from the output to the inverting input ($-$).

                               +------------[ Rf ]------------+
                               |                              |
                               |    |\                        |
                          +----+----| -\                      |
                          |         |   \                     |
                         [R1]  V_in-| +  >--------------------+----o V_out
                          |         |   / 
                         GND        |/    

Mathematical Derivation

1. By virtual ground / virtual short tracking, the inverting terminal voltage tracks the non-inverting terminal: $$V^- = V^+ = V_{in}$$
2. The feedback network forms a resistive voltage divider across $V_{out}$ and ground. Because $I^- = 0\text{ A}$, the current flowing down through $R_1$ equals the current through $R_f$: $$V^- = V_{out} \left( \frac{R_1}{R_1 + R_f} \right)$$
3. Equating $V^-$ to $V_{in}$: $$V_{in} = V_{out} \left( \frac{R_1}{R_1 + R_f} \right)$$
4. Rearranging for voltage gain ($A_v$): $$A_v = \frac{V_{out}}{V_{in}} = \frac{R_1 + R_f}{R_1} = 1 + \frac{R_f}{R_1}$$

Key Non-Inverting Amplifier Characteristics

• Zero Phase Shift: The output is in-phase ($0^\circ$) with the input.
• Gain Magnitude: Voltage gain is always greater than or equal to unity ($A_v \ge 1$).
• Input Impedance ($Z_{in}$): Because the input signal connects directly into the op-amp input gate/base with no shunt resistors, the input impedance seen by the source is exceptionally high ($Z_{in} \approx Z_{in(opamp)} \ge 10^{12}\ \Omega$ for BiFET devices), completely eliminating sensor loading errors.

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The Voltage Follower (Unity-Gain Buffer)

The voltage follower (or unity-gain buffer) is a critical specialized variant of the non-inverting amplifier where the feedback resistor is replaced by a direct short circuit ($R_f = 0\ \Omega$) and the ground resistor is omitted ($R_1 = \infty$).

                               +------------------------------+
                               |                              |
                               |    |\                        |
                               +----| -\                      |
                                    |   \                     |
                               V_in-| +  >--------------------+----o V_out
                                    |   / 
                                    |/    

Applying the non-inverting gain equation:

$$A_v = 1 + \frac{R_f}{R_1} = 1 + \frac{0}{\infty} = 1$$ $$V_{out} = V_{in}$$

Avionics Engineering Significance: Impedance Isolation

While a voltage follower provides zero voltage amplification ($A_v = 1$, or $0\text{ dB}$), it provides immense power gain by performing an impedance transformation:

• Input Impedance: Extremely high ($Z_{in} \approx 10^9\text{ to }10^{12}\ \Omega$).
• Output Impedance: Extremely low ($Z_{out} \approx 0.01\text{ to }0.1\ \Omega$).

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The Summing Amplifier (Inverting Analog Adder)

The summing amplifier provides an analog output proportional to the inverted algebraic sum of multiple input control voltages. Multiple input signals ($V_1, V_2, V_3, \dots$) are connected through individual input resistors ($R_1, R_2, R_3, \dots$) to the op-amp's inverting virtual ground summing junction.

  V1 o----[ R1 ]----+ 
                    |
  V2 o----[ R2 ]----+-----------[ Rf ]------------+
                    |                             |
  V3 o----[ R3 ]----+   |\                        |
                    +---| -\                      |
                        |   \                     |
                   0V --| +  >--------------------+----o V_out
                        |   / 
                        |/    

Mathematical Derivation via KCL

1. At the virtual ground node ($V^- = 0\text{ V}$), each input generates an independent input current: $$I_1 = \frac{V_1}{R_1}, \quad I_2 = \frac{V_2}{R_2}, \quad I_3 = \frac{V_3}{R_3}$$
2. By Kirchhoff's Current Law at the summing node, the total current ($I_{total}$) entering the junction must equal the current flowing across feedback resistor $R_f$: $$I_{total} = I_1 + I_2 + I_3 + \dots = I_f$$ $$\frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3} = -\frac{V_{out}}{R_f}$$
3. Solving for $V_{out}$: $$V_{out} = -\left( \frac{R_f}{R_1}V_1 + \frac{R_f}{R_2}V_2 + \frac{R_f}{R_3}V_3 \right)$$
4. If all input resistors are chosen equal ($R_1 = R_2 = R_3 = R_{in}$), the circuit acts as an equal-gain inverting adder: $$V_{out} = -\frac{R_f}{R_{in}}(V_1 + V_2 + V_3)$$

Aviation Application: Flight Control Surface Mixing

In aircraft fly-by-wire and autopilot systems, summing amplifiers algebraically combine pilot control stick command voltages ($V_{\text{pilot}}$), autopilot pitch displacement commands ($V_{\text{AP}}$), and pitch-rate gyro damping feedback ($V_{\text{gyro}}$) into a single command voltage sent to hydraulic elevator actuator servos.

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The Differential Amplifier (Subtractor)

A differential amplifier amplifies the difference between two input signals while rejecting common-mode electrical interference present on both conductors:

                        +------------[ Rf ]------------+
                        |                              |
  V1 o----[ R1 ]--------+----| -\                      |
                             |   \                     |
  V2 o----[ R2 ]--------+----| +  >--------------------+----o V_out
                        |    |   / 
                       [R3]  |/    
                        |
                       GND

By applying superposition and the voltage divider equation at the non-inverting terminal ($V^+ = V_2 \cdot \frac{R_3}{R_2 + R_3}$):

$$V_{out} = V_2 \left( \frac{R_3}{R_2 + R_3} \right) \left( 1 + \frac{R_f}{R_1} \right) - V_1 \left( \frac{R_f}{R_1} \right)$$

When the resistor ratios are precisely balanced such that $\frac{R_f}{R_1} = \frac{R_3}{R_2}$ (conventionally $R_2 = R_1$ and $R_3 = R_f$):

$$V_{out} = \frac{R_f}{R_1}(V_2 - V_1)$$

The Three-Op-Amp Instrumentation Amplifier

While the single op-amp differential circuit functions well, its input impedance is limited to $R_1$ and $R_2$, which can unbalance bridge circuits. In critical flight-test instrumentation (such as wing strain-gauge bridges and thermocouple exhaust gas temperature sensors), the three-op-amp instrumentation amplifier is standard. It places two non-inverting buffers ahead of the differential stage, providing gigohm input impedances on both lines and allowing voltage gain to be adjusted using a single external gain resistor ($R_G$):

$$A_v = \left( 1 + \frac{2R}{R_G} \right) \left( \frac{R_f}{R_1} \right)$$

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Linear Circuit Topologies Comparison

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Worked Numerical Calculation: Autopilot Elevator Command Mixer

An analog flight-control summing amplifier stage mixes pilot stick pitch command ($V_1$), autopilot pitch correction ($V_2$), and rate-gyro damping ($V_3$). The circuit is energized from dual $\pm 15.0\text{ V DC}$ aircraft avionics power rails, with an output saturation limit of $V_{sat} = \pm 13.5\text{ V}$.

Component parameters:

• Feedback resistor: $R_f = 100\text{ k}\Omega$
• Channel 1 (Pilot): $R_1 = 50\text{ k}\Omega$, $V_1 = +1.20\text{ V}$
• Channel 2 (Autopilot): $R_2 = 100\text{ k}\Omega$, $V_2 = -0.80\text{ V}$
• Channel 3 (Rate Gyro): $R_3 = 25\text{ k}\Omega$, $V_3 = +0.50\text{ V}$

  V1 (+1.2V) o----[ 50k ]-----+ 
                              |
  V2 (-0.8V) o----[ 100k ]----+-----------[ 100k ]----------+
                              |                             |
  V3 (+0.5V) o----[ 25k ]-----+   |\                        |
                              +---| -\                      |
                                  |   \                     |
                             0V --| +  >--------------------+----o V_out
                                  |   / 
                                  |/    

Step 1: Calculate Individual Channel Gains

• Channel 1 Gain: $A_{v1} = -\frac{R_f}{R_1} = -\frac{100\text{ k}\Omega}{50\text{ k}\Omega} = -2.0$
• Channel 2 Gain: $A_{v2} = -\frac{R_f}{R_2} = -\frac{100\text{ k}\Omega}{100\text{ k}\Omega} = -1.0$
• Channel 3 Gain: $A_{v3} = -\frac{R_f}{R_3} = -\frac{100\text{ k}\Omega}{25\text{ k}\Omega} = -4.0$

Step 2: Calculate Individual Output Contributions

• From Channel 1: $V_{out1} = A_{v1} \cdot V_1 = -2.0 \times (+1.20\text{ V}) = -2.40\text{ V}$
• From Channel 2: $V_{out2} = A_{v2} \cdot V_2 = -1.0 \times (-0.80\text{ V}) = +0.80\text{ V}$
• From Channel 3: $V_{out3} = A_{v3} \cdot V_3 = -4.0 \times (+0.50\text{ V}) = -2.00\text{ V}$

Step 3: Compute Total Output Voltage ($V_{out}$)

$$V_{out} = V_{out1} + V_{out2} + V_{out3} = -2.40\text{ V} + 0.80\text{ V} - 2.00\text{ V} = -3.60\text{ V}$$

Step 4: Saturation Check

The resulting output of $-3.60\text{ V}$ is well within the negative saturation rail of $-13.5\text{ V}$. Linear operation is maintained.